$\triangle ABC$ with vertex $A(1,2)$ has equations of internal angle bisectors of $\angle B$ and $\angle C$ as $x-y-1=0$ and $2x+y-9=0$ Respectively. Find the Equation of $BC$
My approach: Solving for $x-y-1=0$ and $2x+y-9=0$ we get the Incentre $I(\frac{10}{3}, \frac{7}{3})$. From here I found that the equation of the angle bisector of $\angle A$ is $x-7y+13=0$.
Please give me a hint on how to procceed.
One way forward would be to look at the angles between the internal angle bisectors. These should be equal to $(\pi/2-\angle B/2-\angle C/2)$, $(\pi/2-\angle A/2-\angle C/2)$ and $(\pi/2-\angle B/2-\angle A/2)$, Solve for the angles from the 3 equations you get, specifically $\angle A$. Getting $\angle A/2$, draw AB and AC from point A using the information. This would also give you BC, once you see where AB, AC intersect the angle bisectors.
There might be a more efficient way, but, I cannot think of any right now. Cheers.